Triangle ABC: Prove cot(A/2) + cot(C/2) = 2cot(B/2)

[anemone]

Here is this week's POTW:

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For a triangle $ABC$ with the sides of $a,\,b,\,$ and $c$ such that $a+c=2b$, prove \(\displaystyle \cot \frac{A}{2}+\cot \frac{C}{2}=2\cot \dfrac{B}{2}\).

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[anemone]

Congratulations to the following members for their correct solution: (Smile)
1. kaliprasad
2. greg1313

Solution from kaliprasad:

Spoiler

By applying the law of cotangents to the triangle $ABC$, with the standard notation $s$ to represent semi-perimeter and the area of triangle as $\triangle$, we have
$\cot(\frac{A}{2}) = \dfrac{s(s-a)}{\triangle}\cdots(1)$
$\cot(\frac{B}{2}) = \dfrac{s(s-b)}{\triangle}\cdots(2)$
$\cot(\frac{C}{2}) = \dfrac{s(s-c)}{\triangle}\cdots(3)$

Add (1) and (3) to get

$\begin{align*}\cot(\frac{A}{2}) + \cot(\frac{C}{2})&=\dfrac{s(s-a)}{\triangle} + \dfrac{s(s-c)}{\triangle}\\&=\dfrac{s(s-a)+s(s-c)}{\triangle}\\&=\dfrac{s(s-a+s-c)}{\triangle}\\&= \dfrac{s(2s-(a+c))}{\triangle}\\&= \dfrac{s(2s-2b)}{\triangle} \text{ (Given)}\\&= 2\dfrac{s(s-b)}{\triangle}\\&=2\cot(\frac{B}{2})\text{ from (2)} \end{align*}$

And we are done.